Here is an example:
(defn f1 [] (lazy-seq (cons 0 (f2))))
(defn f2 [] (lazy-seq (cons 1 (f3))))
(defn f3 [] (lazy-seq (cons 2 (f1))))
In Haskell, the equivalent of the above example would produce a lazy sequence of [0, 1, 2, 0, 1, 2, ...], but in clojure this would lead to a CompilerException because f2 could not be resolved.
Is there any way around this?
use declare to create forward declarations
user> (declare f1)
#'user/f1
user> (declare f2)
#'user/f2
user> (declare f3)
#'user/f3
or as Thumbnail points out:
user> (declare f1 f2 f3)
#'user/f3
works as well
user> (defn f1 [] (lazy-seq (cons 0 (f2))))
#'user/f1
user> (defn f2 [] (lazy-seq (cons 1 (f3))))
#'user/f2
user> (defn f3 [] (lazy-seq (cons 2 (f1))))
#'user/f3
then you get your recursive lazy sequence:
user> (take 20 (f3))
(2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0)
If you just want to produce the lazy sequence, you can define a bunch of mutually recursive functions locally using letfn:
(letfn [(f1 [] (lazy-seq (cons 0 (f2))))
(f2 [] (lazy-seq (cons 1 (f3))))
(f3 [] (lazy-seq (cons 2 (f1))))]
(f1))
=> (0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 ...
Related
i have a list like '(1 2 3 1 4 1 1 6 8 9 0 1) (not actually of numbers, just as an example)
I want to keep all "1" and the element next to the "1".
So the result i would want is (1 2 1 4 1 1 6 1).
Coming from an imperative point of view i would iterate over the list with a for loop, find the "1" at a certain index i and then also keep the element at index i+1.
What would a functional, Clojure idiomatic way of solving this problem be?
Using reduce you can move along the original list building a new list as you go. The reducing function f is passed the new list up until now and the next element from the old list. If the list up until now ends with a 1, or the next element is a 1, add the element to the new list. Otherwise keep the new list as is and move along.
user> (def xs [1 2 3 1 4 1 1 6 8 9 0 1])
#'user/xs
user> (defn f [x y] (if (or (= 1 y) (= 1 (peek x))) (conj x y) x))
#'user/f
user> (reduce f [] xs)
[1 2 1 4 1 1 6 1]
When you can't think of anything clever with sequence combinators, write the recursion by hand. It's not exactly elegant, but it's lazy:
(defn keep-pairs [pred coll]
(lazy-seq
(if (empty? coll)
[]
(let [x (first coll)
xs (next coll)]
(if (pred x)
(cons x (when xs
(let [y (first xs)]
(concat (when-not (pred y) [y])
(keep-pairs pred xs)))))
(when xs
(keep-pairs pred xs)))))))
user> (keep-pairs #{1} [1 2 3 1 4 1 1 6 8 9 0 1])
(1 2 1 4 1 1 6 1)
user> (take 10 (keep-pairs #{1} (cycle [1 2 3])))
(1 2 1 2 1 2 1 2 1 2)
I think I'd prefer reduce for something like this, but here's another 'functional' way of looking at it:
You have a sequence of values that should produce a potentially smaller sequence of values based on some predicate (i.e. filtering) and that predicate needs look-ahead/-behind behavior.
A less common use for map is mapping over multiple sequences at once e.g. (map f coll1 coll2 coll3). If you pass in an "offset" version of the same collection it can be used for the look-ahead/-behind logic.
(defn my-pairs [coll]
(mapcat
(fn [prev curr]
(when (or (= 1 prev) (= 1 curr))
[curr]))
(cons ::none coll) ;; these values are used for look-behind
coll))
This is (ab)using mapcat behavior to combine the mapping/filtering into one step, but it could also be phrased with map + filter.
here's one more solution with clojure's seq processors composition:
(defn process [pred data]
(->> data
(partition-by pred)
(partition-all 2 1)
(filter (comp pred ffirst))
(mapcat #(concat (first %) (take 1 (second %))))))
user> (process #{1} [1 2 1 1 3 4 1 5 1])
;;=> (1 2 1 1 3 1 5 1)
user> (process #{1} [0 1 2 1 1 1 3 4 1 5 1 6])
;;=> (1 2 1 1 1 3 1 5 1 6)
Another idea that does not work since it misses a last one:
(def v [1 2 3 1 4 1 1 6 8 9 0 1])
(mapcat (fn [a b] (when (= a 1) [a b])) v (rest v))
;; => (1 2 1 4 1 1 1 6 1)
So use two arity version of mapcat over the vector and the vector shifted one to the right.
You could check that last 1 explicitly and add, then you get a less elegant working version:
(concat
(mapcat (fn [a b] (when (= a 1) [a b])) v (rest v))
(when (= (peek v) 1) [1]))
;; => (1 2 1 4 1 1 1 6 1)
When you need to loop over data and retain state, I think a plain-old loop/recur is the most straightforward technique:
(ns tst.demo.core
(:use tupelo.core tupelo.test))
(defn keep-pairs
[data]
(loop [result []
prev nil
remaining data]
(if (empty? remaining)
result
(let [curr (first remaining)
keep-curr (or (= 1 curr)
(= 1 prev))
result-next (if keep-curr
(conj result curr)
result)
prev-next curr
remaining-next (rest remaining)]
(recur result-next prev-next remaining-next)))))
(dotest
(let [data [1 2 3 1 4 1 1 6 8 9 0 1]]
(is= [1 2 1 4 1 1 6 1]
(keep-pairs data))))
(defn windowed-pred [n pred]
(let [window (atom [])]
(fn [rf]
(fn ([] (rf))
([acc] (rf acc))
([acc v]
(let [keep? (or (pred v) (some pred #window))]
(swap! window #(vec (take-last n (conj %1 %2))) v)
(if keep?
(rf acc v)
acc)))))))
(let [c [1 2 3 1 4 1 1 6 8 9 0 1]
pred #(= % 1)]
(eduction (windowed-pred 1 pred) c))
(defn last-or-first? [obj pair] (or (= obj (last pair)) (= obj (first pair))))
; to test, whether previous element or element is object
(defn back-shift [l] (cons nil (butlast l))) ;; back-shifts a list
(defn keep-with-follower
[obj l]
(map #'last ; take only the element itself without its previous element
(filter #(last-or-first? obj %) ; is element or previous element the object?
(map #'list (back-shift l) l)))) ; group previous element and element in list
(def l '(1 2 3 1 4 1 1 6 8 9 0 1))
(keep-with-follower 1 l)
;; => (1 2 1 4 1 1 6 1)
A functional solution using only cons first last butlast list map filter = and defn and def.
This question already has an answer here:
Creating the tribonacci numbers in Clojure
(1 answer)
Closed 7 years ago.
Is there a different,cleaner way of generating tribonacci?
(defn tribonacci [x](last (reduce
(fn [[a b c] n]
(conj [b c] (+ a b c))) [0 0 1] (range (- x 3)))))
Here's how I would write it:
(defn tribonacci [a b c]
(lazy-seq (cons a (tribonacci b c (+' a b c)))))
If you want to get the nth tribonacci number, just use, for instance, (tribonacci 0 0 1) and then call nth on that.
you can create a lazy sequence of these numbers, and take the needed element from it:
user> (def t (lazy-cat [1 1 2] (map +' t (rest t) (nthrest t 2))))
#'user/t
user> (take 20 t)
(1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 5768 10609 19513 35890 66012)
replace [1 1 2] to [0 1 1], if you need them starting from 0
I have the following vector, [-1 1 2 -1 3 0 -1 2 -1 4 0 3 0 0]
which represents the tree [[1 2 [3] [2 [4] 3]]]
where -1 begins a new branch and 0 ends it. How can I convert the original vector into a usable tree-like clojure structure (nested vector, nested map)? I think clojure.zip/zipper might do it but I'm not sure how to build those function args.
Zippers are a good tool for this:
(require '[clojure.zip :as zip])
(def in [-1 1 2 -1 3 0 -1 2 -1 4 0 3 0 0])
(def out [[1 2 [3] [2 [4] 3]]])
(defn deepen [steps]
(->> steps
(reduce (fn [loc step]
(case step
-1 (-> loc
(zip/append-child [])
(zip/down)
(zip/rightmost))
0 (zip/up loc)
(zip/append-child loc step)))
(zip/vector-zip []))
(zip/root)))
(assert (= (deepen in) out))
Somehow this feels like cheating:
[(read-string
(clojure.string/join " "
(replace {-1 "[" 0 "]"}
[-1 1 2 -1 3 0 -1 2 -1 4 0 3 0 0])))]
This is not too hard with some recursion:
(defn numbers->tree [xs]
(letfn [(step [xs]
(loop [ret [], remainder xs]
(if (empty? remainder)
[ret remainder]
(let [x (first remainder)]
(case x
0 [ret (next remainder)]
-1 (let [[ret' remainder'] (step (next remainder))]
(recur (conj ret ret'), remainder'))
(recur (conj ret x) (next remainder)))))))]
(first (step xs))))
The idea is to have a function (step) that finds a sub-tree, and returns that tree as well as what numbers are left to be processed. It proceeds iteratively (via loop) for most inputs, and starts a recursive instance of itself when it runs into a -1. The only tricky part is making sure to use the remainder returned from these recursive invocations, rather than proceeding on with the list you were in the middle of.
For given list:
(1 2 3 4)
I'd like to get as output:
(2 3 4 1)
Code I came up with looks like this:
(flatten (cons (rest l) (list (first l))))
However my feeling is, that I overcomplicated this. Any other ideas?
You don't need to flatten a cons, just use concat.
Here is an example:
(let [fruit ["apple" "orange" "grapes" "bananas"]]
(concat (rest fruit) [(first fruit)])
Developing #stonemetal's hint, we can quickly and lazily rotate a vector thus:
(defn rotate [v n]
(let [cv (count v), n (mod n cv)]
(concat (subvec v n cv) (subvec v 0 n))))
It works in either direction:
(map #(rotate (vec (range 5)) %) (range -2 8))
;((3 4 0 1 2)
; (4 0 1 2 3)
; (0 1 2 3 4)
; (1 2 3 4 0)
; (2 3 4 0 1)
; (3 4 0 1 2)
; ...
; (2 3 4 0 1))
So to rotate the first in a sequence to the end:
(rotate (vec (range 1 5)) 1)
You can also use destructuring (either on the function arguments or in a let binding).
(let [[h & tail] '(1 2 3 4)]
(concat tail (list h))) ;=> (1 2 3 4)
If I use the reductions function like so:
(reductions + [1 2 3 4 5])
Then I get
(1 3 6 10 15)
Which is great - but I'd like to apply a binary function in the same way without the state being carried forward - something like
(magic-hof + [1 2 3 4 5])
leads to
(1 3 5 7 9)
ie it returns the operation applied to the first pair, then steps 1 to the next pair.
Can someone tell me the higher-order function I'm looking for? (Something like reductions)
This is my (non-working) go at it:
(defn thisfunc [a b] [(+ a b) b])
(reduce thisfunc [1 2 3 4 5])
You can do it with map:
(map f coll (rest coll))
And if you want a function:
(defn map-pairwise [f coll]
(map f coll (rest coll)))
And if you really need the first element to remain untouched (thanx to juan.facorro's comment):
(defn magic-hof [f [x & xs :as s]]
(cons x (map f s xs)))
partition will group your seq:
user> (->> [1 2 3 4 5] (partition 2 1) (map #(apply + %)) (cons 1))
(1 3 5 7 9)
So, you want to apply a function to subsequent pairs of elements?
(defn pairwise-apply
[f sq]
(when (seq sq)
(->> (map f sq (next sq))
(cons (first sq)))))
Let's try it:
(pairwise-apply + (range 1 6))
;; => (1 3 5 7 9)
This is sufficient:
(#(map + (cons 0 %) %) [1 2 3 4 5])
;; => (1 3 5 7 9)